DC Pandey Solutions: Electromagnetic Induction- 3 | DC Pandey Solutions for JEE Physics PDF Download

 Page 1


24. i i e
t
=
-
0
/ t
Bi i e
T
0 0
=
- / t
t =
æ
è
ç
ö
ø
÷
T
B
ln
1
25. Given, i R P
0
2
= , 
L
R
= t 
when, choke coil is short circuited,
Total heat produced = Magnetic energy
stored in the choke coil
= =
æ
è
ç
ö
ø
÷ =
1
2
1
2
1
2
0
2
Li R
P
R
P ( ) t t
26. i i e
Rt
L
=
-
0
For current to be constant
i i =
0
e
Rt
L
-
= 1
Rt
L
= 0 = not possible.
27. To fi nal time con stant, short the bat tery and
find ef fec tive re sis tance in se ries with
in duc t or
R
R
e
=
2
t = =
L
R
L
R
e
2
28. When switch is at po si tion 1.
In steady state,
i
E
R
1
=
i
2
0 =
When switch is thrown to position 2.
i
E
R
1
= , i
E
R
2
= -
29.
1
2
1
4
1
2
2
0
2
Li L i =
æ
è
ç
ö
ø
÷
i
i
=
0
2
i e
i
t
0
0
1
2
-
æ
è
ç
ç
ö
ø
÷
÷
=
-
t
t = t ln2
Þ              t
L
=
2
2 ln
30. At the mo ment when switch is thrown to
po si tion 2, 
current in capacitor = current in inductor
just before throwing the switch to position 2,
Þ i
E
R
c
=
31. Ini tially, in duc tor of fer s in fi nite re sis tance ,
hence,
i = 0 and 
di
dt
= maximum
\        E V V V
L C R
= + +
But     V V
C R
= =0
Þ      V E
L
=
32. Same as Q.12 ob jec tive Ques tions (Level 2).
33. Let V
0
= Po ten tial of me tal li c rod,
V V B R V B R
B
- = =
0
2
2 2 ( ) w …(i )
V V B R V B R
C 0
2
2 2 - = = ( ) w …(ii)
Adding Eqs. (i) and (ii), we get
V V B R
B C
- = 4
2
w
34. e Blv
c
=
           v
v v
c
=
+
1 2
2
\ e Bl v v = +
1
2
1 2
( )  
or
            e B
dA
dt
=
dA l dx dx = +
1
2
1 2
( )   
    e Bl
dx
dt
dx
dt
= +
æ
è
ç
ö
ø
÷
1
2
1 2
= +
1
2
1 2
Bl v v ( ) 
125 
1
2
i
2
R
L
i
1
R
E
L
R
L
R
Þ
L
R/2
v
2
v
1
dx
1
dx
2
Page 2


24. i i e
t
=
-
0
/ t
Bi i e
T
0 0
=
- / t
t =
æ
è
ç
ö
ø
÷
T
B
ln
1
25. Given, i R P
0
2
= , 
L
R
= t 
when, choke coil is short circuited,
Total heat produced = Magnetic energy
stored in the choke coil
= =
æ
è
ç
ö
ø
÷ =
1
2
1
2
1
2
0
2
Li R
P
R
P ( ) t t
26. i i e
Rt
L
=
-
0
For current to be constant
i i =
0
e
Rt
L
-
= 1
Rt
L
= 0 = not possible.
27. To fi nal time con stant, short the bat tery and
find ef fec tive re sis tance in se ries with
in duc t or
R
R
e
=
2
t = =
L
R
L
R
e
2
28. When switch is at po si tion 1.
In steady state,
i
E
R
1
=
i
2
0 =
When switch is thrown to position 2.
i
E
R
1
= , i
E
R
2
= -
29.
1
2
1
4
1
2
2
0
2
Li L i =
æ
è
ç
ö
ø
÷
i
i
=
0
2
i e
i
t
0
0
1
2
-
æ
è
ç
ç
ö
ø
÷
÷
=
-
t
t = t ln2
Þ              t
L
=
2
2 ln
30. At the mo ment when switch is thrown to
po si tion 2, 
current in capacitor = current in inductor
just before throwing the switch to position 2,
Þ i
E
R
c
=
31. Ini tially, in duc tor of fer s in fi nite re sis tance ,
hence,
i = 0 and 
di
dt
= maximum
\        E V V V
L C R
= + +
But     V V
C R
= =0
Þ      V E
L
=
32. Same as Q.12 ob jec tive Ques tions (Level 2).
33. Let V
0
= Po ten tial of me tal li c rod,
V V B R V B R
B
- = =
0
2
2 2 ( ) w …(i )
V V B R V B R
C 0
2
2 2 - = = ( ) w …(ii)
Adding Eqs. (i) and (ii), we get
V V B R
B C
- = 4
2
w
34. e Blv
c
=
           v
v v
c
=
+
1 2
2
\ e Bl v v = +
1
2
1 2
( )  
or
            e B
dA
dt
=
dA l dx dx = +
1
2
1 2
( )   
    e Bl
dx
dt
dx
dt
= +
æ
è
ç
ö
ø
÷
1
2
1 2
= +
1
2
1 2
Bl v v ( ) 
125 
1
2
i
2
R
L
i
1
R
E
L
R
L
R
Þ
L
R/2
v
2
v
1
dx
1
dx
2
35. Ini tially, ca pac i tor of fer zero re sis tance and
in duc tor of fers in fin i te resistance.
Effective circuit is given by
i
E
R
= = 1 A
36 . i
E
R
e
R t
L
1
1
1
1
= -
æ
è
ç
ç
ö
ø
÷
÷
-
, i
E
R
e
t
R C
2
2
2
=
æ
è
ç
ç
ö
ø
÷
÷
-
       i i i = +
1 2
= -
æ
è
ç
ç
ö
ø
÷
÷
+
-
-
E
R
e
E
R
e
R t
L
t
R C
1 2
1
1
2
at t =
-
10 2
3
ln
 i e e = -
æ
è
ç
ç
ç
ö
ø
÷
÷
÷
+
-
´
´
-
-
-
20
10
1
20
10
10 10 2
10 10
10 2
3
3
3
ln ln
10 10
3
´ ´
-
0.1
   = -
æ
è
ç
ö
ø
÷ +
æ
è
ç
ö
ø
÷ = 2 1
1
2
2
1
2
2 A
37. | |
| |
i
e
R
A
R
dB
dt
= =
= =
- B A
R
B b a
R
0 0
2 2
2 [( ) ] p
=
- B b a
R
0
2 2
4 ( ) p
As inward magnetic field is increasing, net
current must be anticlockwise. Hence
current in inner circle will be clockwise.
38. From Q. 48 Subjective Questions (Level 1).
f = +
æ
è
ç
ö
ø
÷ m
ai a
x
m
p
0
2
1 ln
Case 1
x b = , a a =
f = +
æ
è
ç
ö
ø
÷ m
ai a
b
1
0
2
1
m
p
ln
=
+ æ
è
ç
ö
ø
÷
m
p
0
2
ai b a
b
ln
Case 2
x b a = -
a a =   
    f = +
-
æ
è
ç
ç
ö
ø
÷
÷
m
ai a
b a
2
0
2
1
m
p
ln
        =
-
æ
è
ç
ç
ö
ø
÷
÷
m
p
0
2
ai b
b a
ln
   < > = -
f - f
e
t
m m
2 1
   < > =
< >
= -
f - f
e
e
R Rt
m m
2 1
      q i t
m
R
m
= < > = -
f - f
2
1
    = -
+ æ
è
ç
ö
ø
÷ -
-
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
m
p
0
2
ai
R
b a
b
b
b a
ln ln
       =
-
-
æ
è
ç
ç
ö
ø
÷
÷
m
p
0
2 2
2
ai
R
b
b a
ln
    | | ln q
ai
R
b
b a
=
-
æ
è
ç
ç
ö
ø
÷
÷
m
p
0
2 2
2
39. Mag netic flux linked with the coil.
     f = =
m
nBA
niA
r
m
0
2
    | | e
d
dt
m
=
f
Þ     iR
d
dt
m
=
f
 126
K
E = 20V
5W 4W
6W
C = 0.1mF
5W
ß
C = 0.1mF
R= 10W
2
i
2
i
1
E
R= 10W
1
L = 0.1mH
L = 10 mH
R
R
R
R
E = 5V
ß
E =5V
R/3
R/2
Þ
E =5V
5R/6=5 W
i
R
Page 3


24. i i e
t
=
-
0
/ t
Bi i e
T
0 0
=
- / t
t =
æ
è
ç
ö
ø
÷
T
B
ln
1
25. Given, i R P
0
2
= , 
L
R
= t 
when, choke coil is short circuited,
Total heat produced = Magnetic energy
stored in the choke coil
= =
æ
è
ç
ö
ø
÷ =
1
2
1
2
1
2
0
2
Li R
P
R
P ( ) t t
26. i i e
Rt
L
=
-
0
For current to be constant
i i =
0
e
Rt
L
-
= 1
Rt
L
= 0 = not possible.
27. To fi nal time con stant, short the bat tery and
find ef fec tive re sis tance in se ries with
in duc t or
R
R
e
=
2
t = =
L
R
L
R
e
2
28. When switch is at po si tion 1.
In steady state,
i
E
R
1
=
i
2
0 =
When switch is thrown to position 2.
i
E
R
1
= , i
E
R
2
= -
29.
1
2
1
4
1
2
2
0
2
Li L i =
æ
è
ç
ö
ø
÷
i
i
=
0
2
i e
i
t
0
0
1
2
-
æ
è
ç
ç
ö
ø
÷
÷
=
-
t
t = t ln2
Þ              t
L
=
2
2 ln
30. At the mo ment when switch is thrown to
po si tion 2, 
current in capacitor = current in inductor
just before throwing the switch to position 2,
Þ i
E
R
c
=
31. Ini tially, in duc tor of fer s in fi nite re sis tance ,
hence,
i = 0 and 
di
dt
= maximum
\        E V V V
L C R
= + +
But     V V
C R
= =0
Þ      V E
L
=
32. Same as Q.12 ob jec tive Ques tions (Level 2).
33. Let V
0
= Po ten tial of me tal li c rod,
V V B R V B R
B
- = =
0
2
2 2 ( ) w …(i )
V V B R V B R
C 0
2
2 2 - = = ( ) w …(ii)
Adding Eqs. (i) and (ii), we get
V V B R
B C
- = 4
2
w
34. e Blv
c
=
           v
v v
c
=
+
1 2
2
\ e Bl v v = +
1
2
1 2
( )  
or
            e B
dA
dt
=
dA l dx dx = +
1
2
1 2
( )   
    e Bl
dx
dt
dx
dt
= +
æ
è
ç
ö
ø
÷
1
2
1 2
= +
1
2
1 2
Bl v v ( ) 
125 
1
2
i
2
R
L
i
1
R
E
L
R
L
R
Þ
L
R/2
v
2
v
1
dx
1
dx
2
35. Ini tially, ca pac i tor of fer zero re sis tance and
in duc tor of fers in fin i te resistance.
Effective circuit is given by
i
E
R
= = 1 A
36 . i
E
R
e
R t
L
1
1
1
1
= -
æ
è
ç
ç
ö
ø
÷
÷
-
, i
E
R
e
t
R C
2
2
2
=
æ
è
ç
ç
ö
ø
÷
÷
-
       i i i = +
1 2
= -
æ
è
ç
ç
ö
ø
÷
÷
+
-
-
E
R
e
E
R
e
R t
L
t
R C
1 2
1
1
2
at t =
-
10 2
3
ln
 i e e = -
æ
è
ç
ç
ç
ö
ø
÷
÷
÷
+
-
´
´
-
-
-
20
10
1
20
10
10 10 2
10 10
10 2
3
3
3
ln ln
10 10
3
´ ´
-
0.1
   = -
æ
è
ç
ö
ø
÷ +
æ
è
ç
ö
ø
÷ = 2 1
1
2
2
1
2
2 A
37. | |
| |
i
e
R
A
R
dB
dt
= =
= =
- B A
R
B b a
R
0 0
2 2
2 [( ) ] p
=
- B b a
R
0
2 2
4 ( ) p
As inward magnetic field is increasing, net
current must be anticlockwise. Hence
current in inner circle will be clockwise.
38. From Q. 48 Subjective Questions (Level 1).
f = +
æ
è
ç
ö
ø
÷ m
ai a
x
m
p
0
2
1 ln
Case 1
x b = , a a =
f = +
æ
è
ç
ö
ø
÷ m
ai a
b
1
0
2
1
m
p
ln
=
+ æ
è
ç
ö
ø
÷
m
p
0
2
ai b a
b
ln
Case 2
x b a = -
a a =   
    f = +
-
æ
è
ç
ç
ö
ø
÷
÷
m
ai a
b a
2
0
2
1
m
p
ln
        =
-
æ
è
ç
ç
ö
ø
÷
÷
m
p
0
2
ai b
b a
ln
   < > = -
f - f
e
t
m m
2 1
   < > =
< >
= -
f - f
e
e
R Rt
m m
2 1
      q i t
m
R
m
= < > = -
f - f
2
1
    = -
+ æ
è
ç
ö
ø
÷ -
-
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
m
p
0
2
ai
R
b a
b
b
b a
ln ln
       =
-
-
æ
è
ç
ç
ö
ø
÷
÷
m
p
0
2 2
2
ai
R
b
b a
ln
    | | ln q
ai
R
b
b a
=
-
æ
è
ç
ç
ö
ø
÷
÷
m
p
0
2 2
2
39. Mag netic flux linked with the coil.
     f = =
m
nBA
niA
r
m
0
2
    | | e
d
dt
m
=
f
Þ     iR
d
dt
m
=
f
 126
K
E = 20V
5W 4W
6W
C = 0.1mF
5W
ß
C = 0.1mF
R= 10W
2
i
2
i
1
E
R= 10W
1
L = 0.1mH
L = 10 mH
R
R
R
R
E = 5V
ß
E =5V
R/3
R/2
Þ
E =5V
5R/6=5 W
i
R
  
dq
dt
R
d
dt
m
=
f
 Þ dq
R
d
m
= f
1
     q
nA
rR
di
niA
rR
i
= =
ò
m m
0
0
0
2 2
40. In duced elec tric field in side the re gion of
vary in g mag n eti c fiel ds,
E r
d B
d t
r t x r t x = = + = +
1
2
1
2
6 2 3
2 2
( ) ( ) V/m
At, t = 2.0 s and r
R
= =
2
1.25 cm
      = ´
-
1.25 10
2
 m
           E x = ´ ´ ´ +
-
3 10 4
2
1.25 ( )
=0.3 V/m
         F eE = = ´ ´
-
1.6 0.3 10
19
    = ´
-
48 10
21
 N
41. E r
dB
dt
=
1
2
 Þ E r µ
42. As in ward mag netic field is in creas ing,
in duced elec tric field must be anticlockwise.
43. e
d
dt
a
dB
dt
a B
m
=
f
= = p p
2 2
0
44. E
e
a
aB = =
2
1
2
0
p
45. t a = = qEa i
a = =
´
qEa
ma
q aB a
ma
2
0
2
1
2
           =
qB
m
0
2
46. P t i t = = = × tw t a a ( )
2
      = ´ ma
q B
m
t
2
2
0
2
2
m
At t =1 s
P
q B a
m
=
2
0
2 2
4
47. i
e
R
A
R
dB
dt
= = ×
dB
dt
= 2T/s, A = ´ = 0.2 0.4 0.08 m
2
\    i =
´
´ =
0.08
1.0 1
2 16 A [ Q R r b l = ´ + ( ) 2 ]
As outward magnetic field is increasing,
induced current must be clockwise.
48. e B
dA
dt
A
dB
dt
Blv A
dB
dt
= + = +
At t = 2 s,
B = 4 T, A vt = ´ - = 0.2 0.4 0.06 m ( )
2
    v=5 cm/s = 0.05 m / s
\    e = - ´ ´ + ´ 4 2 0.2 0.05 0.06
     = - + = 0.04 0.12 0.08 V
49. F ilB
e
R
lB = =
=
´
´ ´
0.08
0.8
0.2
1
4
= 0.008 N
50. When ter mi nal ve loc ity is at tained,
power delivered by gravity = power dissipated
in two resistors
       mgv= + 0.76 1.2
          v=
´
=
1.96
0.2 9.8
1 m/s
51. e Blv = = ´ ´ = 0.6 0.6 1 1 V
P
e
R
1
2
1
=   
Þ R
e
P
1
2
1
2
= = =
( ) 0.6
0.76
0.47W
52. P
e
R
2
2
2
= 
Þ R
e
P
2
2
2
2
= = =
( ) 0.6
1.2
0.3W
127 
Page 4


24. i i e
t
=
-
0
/ t
Bi i e
T
0 0
=
- / t
t =
æ
è
ç
ö
ø
÷
T
B
ln
1
25. Given, i R P
0
2
= , 
L
R
= t 
when, choke coil is short circuited,
Total heat produced = Magnetic energy
stored in the choke coil
= =
æ
è
ç
ö
ø
÷ =
1
2
1
2
1
2
0
2
Li R
P
R
P ( ) t t
26. i i e
Rt
L
=
-
0
For current to be constant
i i =
0
e
Rt
L
-
= 1
Rt
L
= 0 = not possible.
27. To fi nal time con stant, short the bat tery and
find ef fec tive re sis tance in se ries with
in duc t or
R
R
e
=
2
t = =
L
R
L
R
e
2
28. When switch is at po si tion 1.
In steady state,
i
E
R
1
=
i
2
0 =
When switch is thrown to position 2.
i
E
R
1
= , i
E
R
2
= -
29.
1
2
1
4
1
2
2
0
2
Li L i =
æ
è
ç
ö
ø
÷
i
i
=
0
2
i e
i
t
0
0
1
2
-
æ
è
ç
ç
ö
ø
÷
÷
=
-
t
t = t ln2
Þ              t
L
=
2
2 ln
30. At the mo ment when switch is thrown to
po si tion 2, 
current in capacitor = current in inductor
just before throwing the switch to position 2,
Þ i
E
R
c
=
31. Ini tially, in duc tor of fer s in fi nite re sis tance ,
hence,
i = 0 and 
di
dt
= maximum
\        E V V V
L C R
= + +
But     V V
C R
= =0
Þ      V E
L
=
32. Same as Q.12 ob jec tive Ques tions (Level 2).
33. Let V
0
= Po ten tial of me tal li c rod,
V V B R V B R
B
- = =
0
2
2 2 ( ) w …(i )
V V B R V B R
C 0
2
2 2 - = = ( ) w …(ii)
Adding Eqs. (i) and (ii), we get
V V B R
B C
- = 4
2
w
34. e Blv
c
=
           v
v v
c
=
+
1 2
2
\ e Bl v v = +
1
2
1 2
( )  
or
            e B
dA
dt
=
dA l dx dx = +
1
2
1 2
( )   
    e Bl
dx
dt
dx
dt
= +
æ
è
ç
ö
ø
÷
1
2
1 2
= +
1
2
1 2
Bl v v ( ) 
125 
1
2
i
2
R
L
i
1
R
E
L
R
L
R
Þ
L
R/2
v
2
v
1
dx
1
dx
2
35. Ini tially, ca pac i tor of fer zero re sis tance and
in duc tor of fers in fin i te resistance.
Effective circuit is given by
i
E
R
= = 1 A
36 . i
E
R
e
R t
L
1
1
1
1
= -
æ
è
ç
ç
ö
ø
÷
÷
-
, i
E
R
e
t
R C
2
2
2
=
æ
è
ç
ç
ö
ø
÷
÷
-
       i i i = +
1 2
= -
æ
è
ç
ç
ö
ø
÷
÷
+
-
-
E
R
e
E
R
e
R t
L
t
R C
1 2
1
1
2
at t =
-
10 2
3
ln
 i e e = -
æ
è
ç
ç
ç
ö
ø
÷
÷
÷
+
-
´
´
-
-
-
20
10
1
20
10
10 10 2
10 10
10 2
3
3
3
ln ln
10 10
3
´ ´
-
0.1
   = -
æ
è
ç
ö
ø
÷ +
æ
è
ç
ö
ø
÷ = 2 1
1
2
2
1
2
2 A
37. | |
| |
i
e
R
A
R
dB
dt
= =
= =
- B A
R
B b a
R
0 0
2 2
2 [( ) ] p
=
- B b a
R
0
2 2
4 ( ) p
As inward magnetic field is increasing, net
current must be anticlockwise. Hence
current in inner circle will be clockwise.
38. From Q. 48 Subjective Questions (Level 1).
f = +
æ
è
ç
ö
ø
÷ m
ai a
x
m
p
0
2
1 ln
Case 1
x b = , a a =
f = +
æ
è
ç
ö
ø
÷ m
ai a
b
1
0
2
1
m
p
ln
=
+ æ
è
ç
ö
ø
÷
m
p
0
2
ai b a
b
ln
Case 2
x b a = -
a a =   
    f = +
-
æ
è
ç
ç
ö
ø
÷
÷
m
ai a
b a
2
0
2
1
m
p
ln
        =
-
æ
è
ç
ç
ö
ø
÷
÷
m
p
0
2
ai b
b a
ln
   < > = -
f - f
e
t
m m
2 1
   < > =
< >
= -
f - f
e
e
R Rt
m m
2 1
      q i t
m
R
m
= < > = -
f - f
2
1
    = -
+ æ
è
ç
ö
ø
÷ -
-
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
m
p
0
2
ai
R
b a
b
b
b a
ln ln
       =
-
-
æ
è
ç
ç
ö
ø
÷
÷
m
p
0
2 2
2
ai
R
b
b a
ln
    | | ln q
ai
R
b
b a
=
-
æ
è
ç
ç
ö
ø
÷
÷
m
p
0
2 2
2
39. Mag netic flux linked with the coil.
     f = =
m
nBA
niA
r
m
0
2
    | | e
d
dt
m
=
f
Þ     iR
d
dt
m
=
f
 126
K
E = 20V
5W 4W
6W
C = 0.1mF
5W
ß
C = 0.1mF
R= 10W
2
i
2
i
1
E
R= 10W
1
L = 0.1mH
L = 10 mH
R
R
R
R
E = 5V
ß
E =5V
R/3
R/2
Þ
E =5V
5R/6=5 W
i
R
  
dq
dt
R
d
dt
m
=
f
 Þ dq
R
d
m
= f
1
     q
nA
rR
di
niA
rR
i
= =
ò
m m
0
0
0
2 2
40. In duced elec tric field in side the re gion of
vary in g mag n eti c fiel ds,
E r
d B
d t
r t x r t x = = + = +
1
2
1
2
6 2 3
2 2
( ) ( ) V/m
At, t = 2.0 s and r
R
= =
2
1.25 cm
      = ´
-
1.25 10
2
 m
           E x = ´ ´ ´ +
-
3 10 4
2
1.25 ( )
=0.3 V/m
         F eE = = ´ ´
-
1.6 0.3 10
19
    = ´
-
48 10
21
 N
41. E r
dB
dt
=
1
2
 Þ E r µ
42. As in ward mag netic field is in creas ing,
in duced elec tric field must be anticlockwise.
43. e
d
dt
a
dB
dt
a B
m
=
f
= = p p
2 2
0
44. E
e
a
aB = =
2
1
2
0
p
45. t a = = qEa i
a = =
´
qEa
ma
q aB a
ma
2
0
2
1
2
           =
qB
m
0
2
46. P t i t = = = × tw t a a ( )
2
      = ´ ma
q B
m
t
2
2
0
2
2
m
At t =1 s
P
q B a
m
=
2
0
2 2
4
47. i
e
R
A
R
dB
dt
= = ×
dB
dt
= 2T/s, A = ´ = 0.2 0.4 0.08 m
2
\    i =
´
´ =
0.08
1.0 1
2 16 A [ Q R r b l = ´ + ( ) 2 ]
As outward magnetic field is increasing,
induced current must be clockwise.
48. e B
dA
dt
A
dB
dt
Blv A
dB
dt
= + = +
At t = 2 s,
B = 4 T, A vt = ´ - = 0.2 0.4 0.06 m ( )
2
    v=5 cm/s = 0.05 m / s
\    e = - ´ ´ + ´ 4 2 0.2 0.05 0.06
     = - + = 0.04 0.12 0.08 V
49. F ilB
e
R
lB = =
=
´
´ ´
0.08
0.8
0.2
1
4
= 0.008 N
50. When ter mi nal ve loc ity is at tained,
power delivered by gravity = power dissipated
in two resistors
       mgv= + 0.76 1.2
          v=
´
=
1.96
0.2 9.8
1 m/s
51. e Blv = = ´ ´ = 0.6 0.6 1 1 V
P
e
R
1
2
1
=   
Þ R
e
P
1
2
1
2
= = =
( ) 0.6
0.76
0.47W
52. P
e
R
2
2
2
= 
Þ R
e
P
2
2
2
2
= = =
( ) 0.6
1.2
0.3W
127 
More than One Cor rect Op tions
1. e B v BLv =
æ
è
ç
ö
ø
÷ =
1
2
1
2
By Fleming’s left hand rule, P must be
positive w.r.t. Q.
2. d BdA Ba dx
m
f = =
=
m
p
0
2
a i
x
dx
f =
m
a i
i
m
p
0
2
2 ln
M
i
a
m
=
f
=
m
p
0
2
2 ln
If the loop is brought close to the wire, upward 
magnetic flux linked with the loop increases,
hence induced current will be clockwise.
3. f = Li = Henry-Am pere.
L
V
di dt
V dt
di
= =
/
 =
Volt-second
Ampere
4. t = =
L
R
1 s
i i e
E
R
e
t t
= - = -
- -
0
1 1 ( ) ( )
/ / t t
= -
-
4 1 ( ) e
t
   
At t = ln2,
            i A =2
Power supplied by battery, P EI = = 16  J/s.
Rate of dissipation of heat in across resistor
= = i R
2
8 J/s
V iR
R
= = 4 V     
       V V E V
a b R
- = - =4 V
5. In both the cases, mag netic flux linked with
in creases, so cur rent i
2
 de creases in or der to
op pose the change.
6. f = = ´ =
1
4 2 8 BA Wb, f =
2
0
    e
t
=
f - f
= =
2 1
8
80
0.1
 V
i
e
R
= = =
80
4
20 A
    q it = = ´ = 20 2 0.1 C
Current is not given as a function of time,
hence heat produced in the coil cannot be
determined.
7. In LC os cil la tions,
w =
1
L C
, f
L C
= =
w
p p 2
1
2
       T
f
LC = =
1
2p
       i q
q
LC
0 0
0
= = w
  
di
dt
q
q
LC
æ
è
ç
ö
ø
÷ = =
max
w
0
0
( )
max
V L
di
dt
q
C
L
=
æ
è
ç
ö
ø
÷ =
max
0
8. If mag netic field in creases, in duced elec tric
field will be anticlockwise and vice-versa.
9. q t = 2
2
i
dq
dt
t = = 4
di
dt
= 4 A/s    
As 
dq
dt
= Positive  
Charge on the capacitor is increasing, hence
current flows from a to b.
t = 1 s, q = 2 C,  i = 4 A    
di
dt
=4 A/s    
      V V L
di
dt
a b
- = = ´ = 1 4 4 V
V V
q
c
b c
- = = =
2
2
1 V  
  V V iR
c d
- = = ´ = 4 4 16 V
  V V
a d
- = + + = 4 1 16 21 V
10. V V Bl
a b
- =
1
2
2
w
V V Bl
c b
- =
1
2
2
w
            V V
a c
- =0
[Direction of velocity of rod a-c is parallel to
length a-c]
 128
dx x
I
a x
1H 4W
d
b
2F
+ – i
a
c
Page 5


24. i i e
t
=
-
0
/ t
Bi i e
T
0 0
=
- / t
t =
æ
è
ç
ö
ø
÷
T
B
ln
1
25. Given, i R P
0
2
= , 
L
R
= t 
when, choke coil is short circuited,
Total heat produced = Magnetic energy
stored in the choke coil
= =
æ
è
ç
ö
ø
÷ =
1
2
1
2
1
2
0
2
Li R
P
R
P ( ) t t
26. i i e
Rt
L
=
-
0
For current to be constant
i i =
0
e
Rt
L
-
= 1
Rt
L
= 0 = not possible.
27. To fi nal time con stant, short the bat tery and
find ef fec tive re sis tance in se ries with
in duc t or
R
R
e
=
2
t = =
L
R
L
R
e
2
28. When switch is at po si tion 1.
In steady state,
i
E
R
1
=
i
2
0 =
When switch is thrown to position 2.
i
E
R
1
= , i
E
R
2
= -
29.
1
2
1
4
1
2
2
0
2
Li L i =
æ
è
ç
ö
ø
÷
i
i
=
0
2
i e
i
t
0
0
1
2
-
æ
è
ç
ç
ö
ø
÷
÷
=
-
t
t = t ln2
Þ              t
L
=
2
2 ln
30. At the mo ment when switch is thrown to
po si tion 2, 
current in capacitor = current in inductor
just before throwing the switch to position 2,
Þ i
E
R
c
=
31. Ini tially, in duc tor of fer s in fi nite re sis tance ,
hence,
i = 0 and 
di
dt
= maximum
\        E V V V
L C R
= + +
But     V V
C R
= =0
Þ      V E
L
=
32. Same as Q.12 ob jec tive Ques tions (Level 2).
33. Let V
0
= Po ten tial of me tal li c rod,
V V B R V B R
B
- = =
0
2
2 2 ( ) w …(i )
V V B R V B R
C 0
2
2 2 - = = ( ) w …(ii)
Adding Eqs. (i) and (ii), we get
V V B R
B C
- = 4
2
w
34. e Blv
c
=
           v
v v
c
=
+
1 2
2
\ e Bl v v = +
1
2
1 2
( )  
or
            e B
dA
dt
=
dA l dx dx = +
1
2
1 2
( )   
    e Bl
dx
dt
dx
dt
= +
æ
è
ç
ö
ø
÷
1
2
1 2
= +
1
2
1 2
Bl v v ( ) 
125 
1
2
i
2
R
L
i
1
R
E
L
R
L
R
Þ
L
R/2
v
2
v
1
dx
1
dx
2
35. Ini tially, ca pac i tor of fer zero re sis tance and
in duc tor of fers in fin i te resistance.
Effective circuit is given by
i
E
R
= = 1 A
36 . i
E
R
e
R t
L
1
1
1
1
= -
æ
è
ç
ç
ö
ø
÷
÷
-
, i
E
R
e
t
R C
2
2
2
=
æ
è
ç
ç
ö
ø
÷
÷
-
       i i i = +
1 2
= -
æ
è
ç
ç
ö
ø
÷
÷
+
-
-
E
R
e
E
R
e
R t
L
t
R C
1 2
1
1
2
at t =
-
10 2
3
ln
 i e e = -
æ
è
ç
ç
ç
ö
ø
÷
÷
÷
+
-
´
´
-
-
-
20
10
1
20
10
10 10 2
10 10
10 2
3
3
3
ln ln
10 10
3
´ ´
-
0.1
   = -
æ
è
ç
ö
ø
÷ +
æ
è
ç
ö
ø
÷ = 2 1
1
2
2
1
2
2 A
37. | |
| |
i
e
R
A
R
dB
dt
= =
= =
- B A
R
B b a
R
0 0
2 2
2 [( ) ] p
=
- B b a
R
0
2 2
4 ( ) p
As inward magnetic field is increasing, net
current must be anticlockwise. Hence
current in inner circle will be clockwise.
38. From Q. 48 Subjective Questions (Level 1).
f = +
æ
è
ç
ö
ø
÷ m
ai a
x
m
p
0
2
1 ln
Case 1
x b = , a a =
f = +
æ
è
ç
ö
ø
÷ m
ai a
b
1
0
2
1
m
p
ln
=
+ æ
è
ç
ö
ø
÷
m
p
0
2
ai b a
b
ln
Case 2
x b a = -
a a =   
    f = +
-
æ
è
ç
ç
ö
ø
÷
÷
m
ai a
b a
2
0
2
1
m
p
ln
        =
-
æ
è
ç
ç
ö
ø
÷
÷
m
p
0
2
ai b
b a
ln
   < > = -
f - f
e
t
m m
2 1
   < > =
< >
= -
f - f
e
e
R Rt
m m
2 1
      q i t
m
R
m
= < > = -
f - f
2
1
    = -
+ æ
è
ç
ö
ø
÷ -
-
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
m
p
0
2
ai
R
b a
b
b
b a
ln ln
       =
-
-
æ
è
ç
ç
ö
ø
÷
÷
m
p
0
2 2
2
ai
R
b
b a
ln
    | | ln q
ai
R
b
b a
=
-
æ
è
ç
ç
ö
ø
÷
÷
m
p
0
2 2
2
39. Mag netic flux linked with the coil.
     f = =
m
nBA
niA
r
m
0
2
    | | e
d
dt
m
=
f
Þ     iR
d
dt
m
=
f
 126
K
E = 20V
5W 4W
6W
C = 0.1mF
5W
ß
C = 0.1mF
R= 10W
2
i
2
i
1
E
R= 10W
1
L = 0.1mH
L = 10 mH
R
R
R
R
E = 5V
ß
E =5V
R/3
R/2
Þ
E =5V
5R/6=5 W
i
R
  
dq
dt
R
d
dt
m
=
f
 Þ dq
R
d
m
= f
1
     q
nA
rR
di
niA
rR
i
= =
ò
m m
0
0
0
2 2
40. In duced elec tric field in side the re gion of
vary in g mag n eti c fiel ds,
E r
d B
d t
r t x r t x = = + = +
1
2
1
2
6 2 3
2 2
( ) ( ) V/m
At, t = 2.0 s and r
R
= =
2
1.25 cm
      = ´
-
1.25 10
2
 m
           E x = ´ ´ ´ +
-
3 10 4
2
1.25 ( )
=0.3 V/m
         F eE = = ´ ´
-
1.6 0.3 10
19
    = ´
-
48 10
21
 N
41. E r
dB
dt
=
1
2
 Þ E r µ
42. As in ward mag netic field is in creas ing,
in duced elec tric field must be anticlockwise.
43. e
d
dt
a
dB
dt
a B
m
=
f
= = p p
2 2
0
44. E
e
a
aB = =
2
1
2
0
p
45. t a = = qEa i
a = =
´
qEa
ma
q aB a
ma
2
0
2
1
2
           =
qB
m
0
2
46. P t i t = = = × tw t a a ( )
2
      = ´ ma
q B
m
t
2
2
0
2
2
m
At t =1 s
P
q B a
m
=
2
0
2 2
4
47. i
e
R
A
R
dB
dt
= = ×
dB
dt
= 2T/s, A = ´ = 0.2 0.4 0.08 m
2
\    i =
´
´ =
0.08
1.0 1
2 16 A [ Q R r b l = ´ + ( ) 2 ]
As outward magnetic field is increasing,
induced current must be clockwise.
48. e B
dA
dt
A
dB
dt
Blv A
dB
dt
= + = +
At t = 2 s,
B = 4 T, A vt = ´ - = 0.2 0.4 0.06 m ( )
2
    v=5 cm/s = 0.05 m / s
\    e = - ´ ´ + ´ 4 2 0.2 0.05 0.06
     = - + = 0.04 0.12 0.08 V
49. F ilB
e
R
lB = =
=
´
´ ´
0.08
0.8
0.2
1
4
= 0.008 N
50. When ter mi nal ve loc ity is at tained,
power delivered by gravity = power dissipated
in two resistors
       mgv= + 0.76 1.2
          v=
´
=
1.96
0.2 9.8
1 m/s
51. e Blv = = ´ ´ = 0.6 0.6 1 1 V
P
e
R
1
2
1
=   
Þ R
e
P
1
2
1
2
= = =
( ) 0.6
0.76
0.47W
52. P
e
R
2
2
2
= 
Þ R
e
P
2
2
2
2
= = =
( ) 0.6
1.2
0.3W
127 
More than One Cor rect Op tions
1. e B v BLv =
æ
è
ç
ö
ø
÷ =
1
2
1
2
By Fleming’s left hand rule, P must be
positive w.r.t. Q.
2. d BdA Ba dx
m
f = =
=
m
p
0
2
a i
x
dx
f =
m
a i
i
m
p
0
2
2 ln
M
i
a
m
=
f
=
m
p
0
2
2 ln
If the loop is brought close to the wire, upward 
magnetic flux linked with the loop increases,
hence induced current will be clockwise.
3. f = Li = Henry-Am pere.
L
V
di dt
V dt
di
= =
/
 =
Volt-second
Ampere
4. t = =
L
R
1 s
i i e
E
R
e
t t
= - = -
- -
0
1 1 ( ) ( )
/ / t t
= -
-
4 1 ( ) e
t
   
At t = ln2,
            i A =2
Power supplied by battery, P EI = = 16  J/s.
Rate of dissipation of heat in across resistor
= = i R
2
8 J/s
V iR
R
= = 4 V     
       V V E V
a b R
- = - =4 V
5. In both the cases, mag netic flux linked with
in creases, so cur rent i
2
 de creases in or der to
op pose the change.
6. f = = ´ =
1
4 2 8 BA Wb, f =
2
0
    e
t
=
f - f
= =
2 1
8
80
0.1
 V
i
e
R
= = =
80
4
20 A
    q it = = ´ = 20 2 0.1 C
Current is not given as a function of time,
hence heat produced in the coil cannot be
determined.
7. In LC os cil la tions,
w =
1
L C
, f
L C
= =
w
p p 2
1
2
       T
f
LC = =
1
2p
       i q
q
LC
0 0
0
= = w
  
di
dt
q
q
LC
æ
è
ç
ö
ø
÷ = =
max
w
0
0
( )
max
V L
di
dt
q
C
L
=
æ
è
ç
ö
ø
÷ =
max
0
8. If mag netic field in creases, in duced elec tric
field will be anticlockwise and vice-versa.
9. q t = 2
2
i
dq
dt
t = = 4
di
dt
= 4 A/s    
As 
dq
dt
= Positive  
Charge on the capacitor is increasing, hence
current flows from a to b.
t = 1 s, q = 2 C,  i = 4 A    
di
dt
=4 A/s    
      V V L
di
dt
a b
- = = ´ = 1 4 4 V
V V
q
c
b c
- = = =
2
2
1 V  
  V V iR
c d
- = = ´ = 4 4 16 V
  V V
a d
- = + + = 4 1 16 21 V
10. V V Bl
a b
- =
1
2
2
w
V V Bl
c b
- =
1
2
2
w
            V V
a c
- =0
[Direction of velocity of rod a-c is parallel to
length a-c]
 128
dx x
I
a x
1H 4W
d
b
2F
+ – i
a
c
Match the Col umns
1.          [ ]
[ ]
[ ][ ]
[ ]
[ ][ ]
B
F
i l
= =
-
MLT
A L
2
   =
- -
[ ] ML T A
0 2 1
         [ ]
[ ][ ]
[ ]
[ ][ ]
[ ]
L
V dt
di
= =
-
ML T T
A
2 3
  =
- -
[ ] ML T A
2 2 2
          [ ] LC =[T ]
2
           [ ] [ ][ ] f =
m
B S
  = =
- - - -
[ [ ] [ ] ML T A ] L ML T A
0 2 1 2 2 2 1
2. i i e
t
= -
-
0
1 ( )
/t
            t = =
L
R
1 s
            i
E
R
0
5 = = A
V iR E e
R
t
= = -
-
( ) 1
V E V Et
L R
t
= - =
-
At t = 0,
V E
L
= = 10 V, V
R
= 0
at          t =1 s
V E e
e
L
= - = -
æ
è
ç
ö
ø
÷
-
( ) 1 1
1
10
1
 V
    V
e
R
=
10
 V
3. In LC oscillations,
w = =
´
=
1 1
1
1
4
2
LC
 rad/s
       q
0
4 = C
       i q
0 0
8 = = w A
  
di
dt
q
æ
è
ç
ö
ø
÷ = =
max
w
2
0
16 A/s.
When, q = 2 C
V V
q
C
L C
= = = 8 V
When, 
di
dt
di
dt
æ
è
ç
ö
ø
÷ =
æ
è
ç
ö
ø
÷ =
1
2
8
max
 A/s.
V V L
di
dt
C L
- = = ´ = 1 8 8 V
4. i
E
R
1
1
9
6
= = = 1.6 A
i
E
R
e e
R t
L
t
2
2
3
1 3 1
2
= - = -
-
-
( ) ( )
/
At t = (ln ) 2 s
V E i R qe
L
t
= - =
-
2 2
3 /
 =
q
2
1 3 /
V i R q e
R
t
2
2 2
3
1 = = -
-
( )
/
 
= -
æ
è
ç
ö
ø
÷ q 1
1
2
1 3 /
V i R
R
1
1 1
9 = = V
V V V
bc L R
= + =
2
9 V
(a ® s), (b ® s), (c ® p), (d ® p).
5. Induced emf
| | e = slope of f - t graph
=
-
-
=
4 0
2 0
2 V
| |
| |
i
e
R
= = =
2
2
1 A
| | | | q i t = = ´ = 1 2 2 C
As current i is constant
H i Rt = = ´ ´ =
2 2
1 2 2 4 ( ) J
129 
R
2
i
2
E
L
R
1
i
1
i
1
2
4
f(Wb)
t